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  刷算法---2021.08
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  <time datetime="2021-08-12T08:20:46.000Z" itemprop="datePublished">2021-08-12</time>
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  <blockquote>
<p>本文将记录我在2021年8月中所刷过的题目，题目量不多，但是每一道保证理解透彻。</p>
</blockquote>
<h1 id="力扣26题：删除有序数组中的重复项"><a href="#力扣26题：删除有序数组中的重复项" class="headerlink" title="力扣26题：删除有序数组中的重复项"></a><a target="_blank" rel="noopener" href="https://leetcode-cn.com/problems/remove-duplicates-from-sorted-array/">力扣26题：删除有序数组中的重复项</a></h1><p>首先看清楚题目，这道题目说了是一个有序的数组，那么重复的元素必定是连续的，就连续这一特点我们可以摒弃平常用的字典计数的方法或者使用set的方法，而采用更加简便的方法。</p>
<ul>
<li><font color=gree>思路1：双指针(时间复杂度O(n)，空间复杂度O(1))</font></li>
</ul>
<p>假设两个指针p和q，p在左边，q在右边，如果$nums[p]==nums[q]$，即两指针指向的元素值相等了，那么q指针右移一个单位，再次判断俩指针指向的值是否相等，如果还是相等，q指针继续移动；如果不相等，p指针移动一个单位，且$nums[p] = nums[q]$。</p>
<p>那么何时跳出循环呢？如果q到达了nums最后一个元素，那么说明了所有的元素都遍历了一遍，那么此时p所指向的就是不重复的元素的最后一个。</p>
<span id="more"></span>

<p>将上述过程写成伪代码：</p>
<figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br></pre></td><td class="code"><pre><span class="line">p=<span class="number">0</span>, q=<span class="number">1</span>, r=<span class="number">0</span></span><br><span class="line"><span class="keyword">while</span> q &lt; nums.size():</span><br><span class="line">    <span class="keyword">if</span> nums[p] == nums[q]:</span><br><span class="line">        q++</span><br><span class="line">    <span class="keyword">else</span>:</span><br><span class="line">        p++</span><br><span class="line">        nums[p] = nums[q]</span><br><span class="line"><span class="keyword">return</span> p</span><br><span class="line"></span><br><span class="line"><span class="comment"># 这里需要注意的是，如果直接返回p结果是错的，因为测试的方式是p在nums中所对应的元素是不访问的，</span></span><br><span class="line"><span class="comment"># 所以需要返回p+1</span></span><br></pre></td></tr></table></figure>



<p>可以自行举个例子来将上述过程推演一遍：<br>nums = 0, 0, 1, 1, 1, 2, 2, 3, 3, 4。</p>
<p>完整代码：</p>
<figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br></pre></td><td class="code"><pre><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span> &#123;</span></span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function"><span class="keyword">int</span> <span class="title">removeDuplicates</span><span class="params">(vector&lt;<span class="keyword">int</span>&gt;&amp; nums)</span> </span>&#123;</span><br><span class="line">        <span class="keyword">if</span> (nums.<span class="built_in">empty</span>()) <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">        </span><br><span class="line">        <span class="keyword">int</span> p = <span class="number">0</span>, q = <span class="number">1</span>;</span><br><span class="line">        <span class="keyword">while</span> (q &lt; nums.<span class="built_in">size</span>()) &#123;</span><br><span class="line">            <span class="keyword">if</span> (nums[p] == nums[q]) &#123;</span><br><span class="line">                q++;</span><br><span class="line">            &#125;</span><br><span class="line">            <span class="keyword">else</span> &#123;</span><br><span class="line">                p++;</span><br><span class="line">                nums[p] = nums[q];</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> p + <span class="number">1</span>;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br><span class="line"></span><br><span class="line"><span class="comment">// 代码看起来不够简洁？将一些语句合并之后得到如下的版本：</span></span><br><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span> &#123;</span></span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function"><span class="keyword">int</span> <span class="title">removeDuplicates</span><span class="params">(vector&lt;<span class="keyword">int</span>&gt;&amp; nums)</span> </span>&#123;</span><br><span class="line">        <span class="keyword">if</span> (nums.<span class="built_in">empty</span>()) <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">        </span><br><span class="line">        <span class="keyword">int</span> p = <span class="number">0</span>, q = <span class="number">1</span>;</span><br><span class="line">        <span class="keyword">for</span> (<span class="keyword">int</span> q = <span class="number">1</span>; q &lt; nums.<span class="built_in">size</span>(); q++) &#123;</span><br><span class="line">            <span class="keyword">if</span> (nums[p] != nums[q]) </span><br><span class="line">                nums[++p] = nums[q];</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> p + <span class="number">1</span>;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<ul>
<li><font color=gree>思路2：使用STL(时间复杂度O(n)，空间复杂度O(1))</font></li>
</ul>
<figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br></pre></td><td class="code"><pre><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span> &#123;</span></span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function"><span class="keyword">int</span> <span class="title">removeDuplicates</span><span class="params">(vector&lt;<span class="keyword">int</span>&gt;&amp; nums)</span> </span>&#123;</span><br><span class="line">        <span class="keyword">return</span> <span class="built_in">distance</span>(nums.<span class="built_in">begin</span>(), </span><br><span class="line">        <span class="built_in">removeDuplicates</span>(nums.<span class="built_in">begin</span>(), nums.<span class="built_in">end</span>(), nums.<span class="built_in">begin</span>()));</span><br><span class="line">    &#125;</span><br><span class="line"></span><br><span class="line">    <span class="function"><span class="keyword">template</span>&lt;<span class="keyword">typename</span>  InIt, <span class="keyword">typename</span> OutIt&gt;</span></span><br><span class="line"><span class="function">    OutIt <span class="title">removeDuplicates</span><span class="params">(InIt first, InIt last, OutIt output)</span> </span>&#123;</span><br><span class="line">        <span class="keyword">while</span> (first != last) &#123;</span><br><span class="line">            *output++ = *first;</span><br><span class="line">            first = <span class="built_in">upper_bound</span>(first, last, *first);</span><br><span class="line">        &#125;</span><br><span class="line"></span><br><span class="line">        <span class="keyword">return</span> output;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<ul>
<li><font color=gree>思路3：通用解法，将原问题的「最多保留 1 位」修改为「最多保留 k 位」。</font></li>
</ul>
<p>假设我们使用的$nums=[1, 1, 1, 2, 2, 3]$<br>我们仍然使用双指针法。这样的思路：</p>
<p>既然需要保留K个相同的数字，我们就用一个变量r来存储当前数字的相同的个数。一开始p指针指向index=0，q指针指向index=1，首先判断$nums[p]==nums[q]$是否成立，如果是的，我们需要判断计数器r的值是否超出了K的限定范围，如果超出了，那么只用q继续往后走，如果没有超出，我们让与p相距r个位置的数值=nums[q]，即：$nums[p+r] = nums[q]$；如果不成立，即此时$nums[p]!=nums[q]$，假设刚刚在遍历数字1，那么此时意味着此时数字1已经遍历完了，所以$nums[q]=2$，这个时候我们要将p指针指向这个新的数字的开端，即p=q。</p>
<p>这样写有什么问题？</p>
<p>如果p和q之间相差很多呢，比如1， 1， 1， 1， 1， 1，2，此时p指向第三个1，而q指向2，能将p直接指向q的2吗？显然不能！但是想起来我们用了一个计数器r，这就容易想到我们可以使用计数器来实现p指针的新指向：$p += r$。然后是一个新的数字了，所以计数器r需要更新为1，新的数字还要移动过来给p：$nums[p] = nums[q]$，这就是一个完整的循环需要做的事情。循环结束的标志？和K为1的时候一样，当q超出nums长度即停止。</p>
<p>综合上述流程的伪代码为：</p>
<figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br></pre></td><td class="code"><pre><span class="line">k=<span class="number">2</span>, p = <span class="number">0</span>, q = <span class="number">1</span>, r = <span class="number">1</span></span><br><span class="line"><span class="keyword">while</span> q &lt; nums.size():</span><br><span class="line">	<span class="keyword">if</span> nums[p] == nums[q]:</span><br><span class="line">		nums[p + r] = nums[q]</span><br><span class="line">		r++, q++</span><br><span class="line">	<span class="keyword">else</span>:</span><br><span class="line">		p += r, r = <span class="number">1</span></span><br><span class="line">		nums[p] = nums[q]</span><br><span class="line">		q++</span><br></pre></td></tr></table></figure>

<p>使用C++来实现：</p>
<figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br></pre></td><td class="code"><pre><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span> &#123;</span></span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function"><span class="keyword">int</span> <span class="title">removeDuplicates</span><span class="params">(vector&lt;<span class="keyword">int</span>&gt;&amp; nums)</span> </span>&#123;</span><br><span class="line">        <span class="keyword">if</span> (nums.<span class="built_in">empty</span>()) <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line"></span><br><span class="line">        <span class="keyword">int</span> k = <span class="number">2</span>, r = <span class="number">1</span>, p = <span class="number">0</span>, q = <span class="number">1</span>;</span><br><span class="line"></span><br><span class="line">        <span class="keyword">for</span> (; q &lt; nums.<span class="built_in">size</span>(); q++) &#123;</span><br><span class="line">           <span class="keyword">if</span> (nums[p] == nums[q]) &#123;</span><br><span class="line">               <span class="keyword">if</span> (r &lt; k) &#123;</span><br><span class="line">                   nums[p + r] = nums[q];</span><br><span class="line">                   r++;</span><br><span class="line">               &#125;</span><br><span class="line">           &#125;</span><br><span class="line">           <span class="keyword">else</span> &#123;</span><br><span class="line">               p += r;</span><br><span class="line">               r = <span class="number">1</span>;</span><br><span class="line">               nums[p] = nums[q];</span><br><span class="line">           &#125;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> p + <span class="number">1</span>;</span><br><span class="line">        &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<p>问题来了，将这段C++代码拿去运行发现只能通过部分案例，什么问题？仔细分析代码之后发现，如果在退出循环之前，进入的是else分支，那么p将得到更新：$p+=r$，如果经过的是if分支，p无法得到更新，直接退出循环！所以这里逻辑需要更改，可以直接在if分支中让每一次p+r吗?不行，因为r是不断增加的。我的解决方式是，每一次都让p更新一下，而不是到了最后一次性加上r，即在每一次遇到与nums[p]相同数字的时候，如果r&lt;k就马上更新p：</p>
<figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">if</span> (nums[p] == nums[q]) &#123;</span><br><span class="line">    <span class="keyword">if</span> (r &lt; k) &#123;</span><br><span class="line">        r++;</span><br><span class="line">        nums[++p] = nums[q];</span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>

<p>这样就可以将else分支中的p+=r去掉：</p>
<figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">else</span> &#123;</span><br><span class="line"><span class="comment">//  p += r;</span></span><br><span class="line">    r = <span class="number">1</span>;</span><br><span class="line">    nums[++p] = nums[q];</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>

<p>完整的代码为：</p>
<figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br></pre></td><td class="code"><pre><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span> &#123;</span></span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function"><span class="keyword">int</span> <span class="title">removeDuplicates</span><span class="params">(vector&lt;<span class="keyword">int</span>&gt;&amp; nums)</span> </span>&#123;</span><br><span class="line">        <span class="keyword">if</span> (nums.<span class="built_in">empty</span>()) <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line"></span><br><span class="line">        <span class="keyword">int</span> k = <span class="number">2</span>, r = <span class="number">1</span>, p = <span class="number">0</span>, q = <span class="number">1</span>;</span><br><span class="line"></span><br><span class="line">        <span class="keyword">for</span> (; q &lt; nums.<span class="built_in">size</span>(); q++) &#123;</span><br><span class="line">           <span class="keyword">if</span> (nums[p] == nums[q]) &#123;</span><br><span class="line">               <span class="keyword">if</span> (r &lt; k) &#123;</span><br><span class="line">                   r++;</span><br><span class="line">                   nums[++p] = nums[q];</span><br><span class="line">               &#125;</span><br><span class="line">           &#125;</span><br><span class="line">           <span class="keyword">else</span> &#123;</span><br><span class="line"><span class="comment">//               p += r;</span></span><br><span class="line">               r = <span class="number">1</span>;</span><br><span class="line">               nums[++p] = nums[q];</span><br><span class="line">           &#125;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> p + <span class="number">1</span>;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<p>这个搞定之后，可以顺便去把这道题给AC掉：<br><a target="_blank" rel="noopener" href="https://leetcode-cn.com/problems/remove-duplicates-from-sorted-array-ii/">80. 删除有序数组中的重复项 II</a></p>
<h1 id="力扣33题-搜索旋转排序数组"><a href="#力扣33题-搜索旋转排序数组" class="headerlink" title="力扣33题.搜索旋转排序数组"></a><a target="_blank" rel="noopener" href="https://leetcode-cn.com/problems/search-in-rotated-sorted-array/">力扣33题.搜索旋转排序数组</a></h1><p>这道题如果不去考虑时间复杂度的限制将是非常容易的遍历问题，但是本题的进阶方案是：设计一个时间复杂度为$O(log n)$的解决方案。</p>
 
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